another integral w/ sub u

I am trying not to use formulas..

$\int_{}{}\ (^4\sqrt{3x+5}) (x) dx$

Answer : $\frac{4}{81} (3x+5)^{\frac{5}{4} }(3x-4)+C$

I am pretty sure I am on the right track but my weak algebra is making it harder

u= 3x+5
u'= 3

dx in tems of u = $\frac {1}{3}$
and
x in tems of the new intgral = $(\frac{u-5}{3})$

Here is the new integral!!!!!! I am going to simplify it now!!!!!!!
$\frac {1}{3} \int_{}{} (\frac{u-5}{3}) u^{ \frac {1}{4} } du$

Factoring out the 1/3
$\frac {1}{3} \int_{}{} (\frac{1}{3})(u-5) u^{ \frac {1}{4} } du$

$\frac {1}{9} \int_{}{} (u-5) u^{ \frac {1}{4} } du$

distribute the $u^{1/4}$, Notice that I can now directly integrate the function.

$\frac {1}{9} ( \int_{}{} u ^{ \frac {5}{4} }du -5 \int_{}{} u ^{ \frac {1}{4} }du)$

$\frac {1}{9} [ \frac {4}{9} u ^{ \frac {9}{4} }- 4u ^{ \frac {5}{4} } ]$

factor out common crap

$\frac {1}{9} [ \frac {4}{9} u ^{ \frac {5}{4} }(u- 9 )]$ = $( \frac {4}{81} u ^{ \frac {5}{4} })(u- 9 )$

plug U back in

= $( \frac {4}{81} (3x+5) ^{ \frac {5}{4} })(3x+5- 9 )$

simplify and dont forget to add C!!!!!!!!!!

= $\frac {4}{81} (3x+5) ^{ \frac {5}{4} }(3x-4 )+C$